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The union of a chain of dominated extensions is a well-defined dominated linear functional

Statement

Let $X$ be a real vector space and let $p : X \to R$ be sublinear. Let $C$ be a nonempty chain, ordered by extension, of pairs $(N, g)$ such that $N \subseteq X$ is a linear subspace and $g : N \to R$ is linear with $g (n) \leq p (n)$ for every $n \in N$ .

Put

$U : = ⋃_{(N, g) \in C} N .$

Then $U$ is a linear subspace of $X$ , and the pointwise union $G : U \to R$ defined by $G (x) = g (x)$ whenever $x \in N$ is a well-defined linear functional with $G (u) \leq p (u)$ for every $u \in U$ .

If every $(N, g) \in C$ extends the same linear functional $f : M \to R$ , then $G$ also extends $f$ .

Facts & Assumptions

Given: A real vector space $X$ , a sublinear functional $p : X \to R$ , and a nonempty chain $C$ of dominated linear functionals ordered by extension.

[L1]

A linear functional is additive and homogeneous over the scalar field (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ).

[L2]

A chain is a subset in which any two elements are comparable (Chain in a poset).

Proof

technique · direct

1.1

If $x \in N_{1} \cap N_{2}$ for $(N_{1}, g_{1}), (N_{2}, g_{2}) \in C$ , then [L2] gives comparability. Suppose $N_{1} \subseteq N_{2}$ . Since the chain order is extension, $g_{2} ∣_{N_{1}} = g_{1}$ , so $g_{1} (x) = g_{2} (x)$ . The other inclusion case is the same. Therefore $G (x)$ is well defined on overlaps.

L2givenconstruct

1.2

Let $u, v \in U$ and $a \in R$ . Choose $(N_{1}, g_{1}), (N_{2}, g_{2}) \in C$ with $u \in N_{1}$ and $v \in N_{2}$ . By [L2], one of the domains contains the other; after relabeling, assume $N_{1} \subseteq N_{2}$ . Then $u, v \in N_{2}$ , so $u + v \in N_{2}$ and $a u \in N_{2}$ because $N_{2}$ is a linear subspace. Hence $u + v, a u \in U$ . Thus $U$ is a linear subspace.

L2givenalgebra

2.1

With the same choice of $N_{2}$ as in step 1.2, one has $G (u + v) = g_{2} (u + v) = g_{2} (u) + g_{2} (v) = G (u) + G (v),$ and $G (a u) = g_{2} (a u) = a g_{2} (u) = a G (u),$ by [L1]. Therefore $G$ is linear.

step 1.1step 1.2L1

3.1

If $u \in U$ , choose $(N, g) \in C$ with $u \in N$ . Then $G (u) = g (u) \leq p (u)$ by the defining property of the chain element, so $G$ is dominated by $p$ . If every chain element extends the same $f : M \to R$ , then every $m \in M$ lies in each domain and all values there equal $f (m)$ , so $G ∣_{M} = f$ .

step 1.1given∎

Depends on

Used by

Hahn-Banach dominated extension theorem for real vector spaces Theorem

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Daniel Daners, Introduction to Functional Analysis, Theorem 26.1 (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis, Theorem 4.13 (standard reference, not scraped)